Accelerating Clock

After $17$ minutes the second hand will have made

$17 + \dfrac{1}{12}\displaystyle\sum_{n=1}^{17} n = 17 + \dfrac{1}{12} \times \dfrac{17*18}{2} = 17 + \dfrac{153}{12} = 29 + \dfrac{9}{12}$ revolutions,

where the identity $\displaystyle\sum_{n=1}^{N} n = \dfrac{N(N + 1)}{2}$ was used. (Reference: sum of n, n², or n³ )

This "extra" $\dfrac{9}{12}$ th of a revolution results in the second hand pointing at the number $\boxed{9}$ .

Essentially, after $N$ minutes the number the second hand will be pointing at is $\displaystyle\left(\sum_{n=1}^{N} n\right) \pmod{12}$ .

For the bonus question, we note that since there are $24*60$ minutes in a day, and since

$\displaystyle\sum_{n=1}^{24*60} n = \dfrac{(24*60) \times (24*60 + 1)}{2} = (12*60)*(24*60 + 1)$

is divisible by $12$ , the second hand will make an integral number of revolutions, leaving the second hand pointing back at $\boxed{12}$ after a day.

constructing the revolution period into a series - 1 $\frac{1}{12}$ , 1 $\frac{2}{12}$ , 1 $\frac{3}{12}$ and after 17 min. last rev. is= 2 $\frac{5}{12}$ => $\frac{13}{12}$ + $\frac{14}{12}$ + $\frac{15}{12}$ ..... $\frac{29}{12}$ . Where- first term, a= $\frac{13}{12}$ and last term, n = $\frac{29}{12}$ .

So, total revolution of the second hand in 17 minutes is S= \frac{17( \(\frac{13}{12} + $\frac{29}{12}$ )}{2}) = $\frac{357}{12}$ . and dividing 357 by 12 gives us 9 as remainder and 29 as quotient. What means that 2nd hand will make 29 perfect revolution and then will stuck to 9 after 17 minutes. !